A champagne bottle is held upright 1.2 m above the floor
as the wire around its cork is removed. The cork then pops out, rises
vertically, and falls to the floor 1.4 s later.

What was the cork's initial velocity?

a)

3.2 m/s

b)

4.9 m/s

c)

7.3 m/s

N

d)

6.0 m/s

e)

None of the above.

What height above the bottle did the cork reach?

a)

2.12 m

b)

1.43 m

c)

2.21 m

N

d)

1.84 m

e)

None of the above.

Solution:

Solution:

For solving this problem,
use can be made of the time needed by the cork to reach the floor,
t = 1.4 s, which we know is
1.2 m below the launching point of the cork. This quantity correspond to
the cork displacement of y = -1.2 mhe initial velocity of the cork can be calculated
using the formula g is the
acceleration of gravity. Thus, solving for the initial velocity,v0y,in the previous relation, .
And the initial velocity becomes
.

Knowing the initial velocity
of the cork, the maximum height can be calculated from the formula,
,
(F_F 5), since
at
the velocity of the cork is zero,
.
The maximum height can be calculated,
.
Substituting the numerical values
.

its final velocity?

a)

- 6.5 m/s

N

b)

- 7.7 m/s

c)

- 8.5 m/s

d)

- 5.4 m/s

e)

None of the above.

Solution:

At the floor level, the final velocity of
the cork is obtained directly from the formula
that evaluated at the known numerical values results in
.